[leetcode]Pow(x, n)

a^n = a^(n/2) * a(n/2)...二分

 

class Solution {
public:
    double epow(double x,int n){
        if(n == 0) return 1;
        double tmp = epow(x , n / 2);
        if(n % 2 == 1)
            return tmp * tmp * x;
        else return tmp * tmp;
    }
    double pow(double x, int n) {
        if(abs(x - 0) < 1e-10) return 0;
        if(n < 0) return 1/epow(x,-n);else
        return epow(x,n);
    }
};